Post on 16-Jan-2016
Copyright © 2011 Pearson Education, Inc.
Rational Expressions and Equations
CHAPTER
7.1 Simplifying Rational Expressions7.2 Multiplying and Dividing Rational Expressions7.3 Adding and Subtracting Rational Expressions with the
Same Denominator7.4 Adding and Subtracting Rational Expressions with
Different Denominators7.5 Complex Rational Expressions7.6 Solving Equations Containing Rational Expressions7.7 Applications with Rational Expressions, Including
Variation
77
Copyright © 2011 Pearson Education, Inc.
Simplifying Rational Expressions7.17.1
1. Evaluate rational expressions.2. Find numbers that cause a rational expression to be
undefined.3. Simplify rational expressions containing only monomials.4. Simplify rational expressions containing multiterm
polynomials.
Slide 7- 3Copyright © 2011 Pearson Education, Inc.
Rational expression: An expression that can be
written in the form , where P and Q are
polynomials and Q 0.
P
Q
Some rational expressions are
5
2
3
18
x
x 2
7
9
x
x
2 2 15
4 20
x x
x
Slide 7- 4Copyright © 2011 Pearson Education, Inc.
Example 1
Evaluate the expression when
a. x = 2 b. x = –1
7 9,
1
x
x
Solution
a. 7 9
1
x
x
7
1
2
2
9
14 9
3
5 2 or 1
3 3
b. 7 9
1
x
x
7 91
1 1
7 9
0
16
0
which is undefined
Slide 7- 5Copyright © 2011 Pearson Education, Inc.
Finding Values That Make a Rational Expression UndefinedTo determine the value(s) that make a rational expression undefined, 1. Write an equation that has the denominator set equal to zero. 2. Solve the equation.
Slide 7- 6Copyright © 2011 Pearson Education, Inc.
Example 2aFind every value for the variable that makes the expression undefined.4
6 5
z
z
Solution 6 5 0z 6 5z
The original expression is undefined if z is replaced by 5/6.
Set the denominator equal to 0.
Add 5 to both sides.
Divide both sides by 6.5
6z
Slide 7- 7Copyright © 2011 Pearson Education, Inc.
Example 2bFind every value for the variable that makes the expression undefined.
3 2
6
5 4y y y
Solution 3 25 4 0y y y
2 5 4 0y y y 4 1 0y y y
0 or 4 0 or 1 0y y y 4 1y y
The original expression is undefined if y is replaced by 0, –4, or –1.
Set the denominator equal to 0.
Factor out the monomial GCF, y.
Use the zero factor theorem.
Slide 7- 8Copyright © 2011 Pearson Education, Inc.
Simplifying Rational ExpressionsTo simplify a rational expression to lowest terms:
1. Factor the numerator and denominator completely. 2. Divide out all the common factors in the numerator
and denominator.3. Multiply the remaining factors in the numerator and
the remaining factors in the denominator.
1, where , , and
1
are polynomials and and are not 0.
PR P PP Q
QR Q Q
R Q R
Slide 7- 9Copyright © 2011 Pearson Education, Inc.
Example 3
Simplify
Solution
7
3
4.
24
x
x
2 2
2 2 2 3
x x x x x x x
x x x
4
6
x
Write the numerator and denominator in factored form, then eliminate the common factors. Multiply the remaining factors.
7
3
4
24
x
x
Slide 7- 10Copyright © 2011 Pearson Education, Inc.
Example 4 Simplify.a. b.
Solution
a.
b.
3
5
6
42
a
a
2 2
2 4
24
72
xy z
x y
3
5
6
42
a
a
2 3
2 3 7
a a a
a a a a a
2
1
7a
2 2
2 4
24
72
xy z
x y
2 2 2 3
2 2 2 3 3
x y y z z
x x y y y y
2
23
z
xy
Write the numerator and denominator in factored form, then eliminate the common factors. Multiply the remaining factors.
Slide 7- 11Copyright © 2011 Pearson Education, Inc.
Example 5
Simplify .
8
2 4
xy
x
Solution
2 2 2
2 2
x y
x
Write the numerator and denominator in factored form, then divide out the common factor, which is 2.
8
2 4
xy
x
4
2
xy
x
Slide 7- 12Copyright © 2011 Pearson Education, Inc.
Example 6a
Simplify.
2
2
6 12
7 14
x x
x x
Solution
6 2
7 2
x x
x x
6 2
7 2
x x
x x
6
7
Write the numerator and denominator in factored form, then divide out the common factors, x and x + 2.
2
2
6 12
7 14
x x
x x
Slide 7- 13Copyright © 2011 Pearson Education, Inc.
Example 6b
Simplify.
2
2
25
2 15
y
y y
Solution
5 5
3 5
y y
y y
Factor the numerator and denominator completely. Then divide out the common factor, y – 5.
2
2
25
2 15
y
y y
5
3
y
y
Slide 7- 14Copyright © 2011 Pearson Education, Inc.
Example 6c
Simplify.
2
2
3 9 12
6 30 24
x x
x x
SolutionFactor out the GCF.
2
2
3 9 12
6 30 24
x x
x x
2
2
3 3 4
6 5 4
x x
x x
3 4 1
6 4 1
x x
x x
3 4 1
2 3 4 1
x x
x x
4
2 4
x
x
Factor the polynomial factors.
Divide out common factors.
Slide 7- 15Copyright © 2011 Pearson Education, Inc.
Example 7
Simplify.
2
2
4 4 48
3 3 18
x x
x x
Solution
4 4 4 16 or
3 2 3 6
x x
x x
4 4 3
3 2 3
x x
x x
2
2
4 4 48
3 3 18
x x
x x
Slide 7- 16Copyright © 2011 Pearson Education, Inc.
Example 8
Simplify.
22 21
6 2
x x
x
Solution
2 7 3
2 3
x x
x
22 21
6 2
x x
x
2 7 1 3
2 3
x x
x
2 7 2 7 or
2 2
x x
Slide 7- 17Copyright © 2011 Pearson Education, Inc.
Simplify.
a) 5
b)
c)
d)
2
2
6 4
2 2
n n
n n
2
2
3 2n n
n n
23 2
1
n n
n
3 2
1
n
n
7.1
Slide 7- 18Copyright © 2011 Pearson Education, Inc.
Simplify.
a) 5
b)
c)
d)
2
2
6 4
2 2
n n
n n
2
2
3 2n n
n n
23 2
1
n n
n
3 2
1
n
n
7.1
Slide 7- 19Copyright © 2011 Pearson Education, Inc.
Simplify.
a) –10x
b)
c)
d)
2
2
10 25
25
x x
x
10
5
x
x
5
5
x
x
5
5
x
x
7.1
Slide 7- 20Copyright © 2011 Pearson Education, Inc.
Simplify.
a) –10x
b)
c)
d)
2
2
10 25
25
x x
x
10
5
x
x
5
5
x
x
5
5
x
x
7.1
Copyright © 2011 Pearson Education, Inc.
Multiplying and Dividing Rational Expressions7.27.2
1. Multiply rational expressions.2. Divide rational expressions.3. Convert units of measurement using dimensional
analysis.
Slide 7- 22Copyright © 2011 Pearson Education, Inc.
, where , , , P R PR
P Q RQ S QS
Multiplying Rational ExpressionsTo multiply rational expressions,
1. Factor each numerator and denominator completely.
2. Divide out any numerator factor with any matching denominator factor.
3. Multiply numerator by numerator and denominator by denominator.
4. Simplify as needed.
and are polynomials and 0 and 0.S Q S
Slide 7- 23Copyright © 2011 Pearson Education, Inc.
Example 1
Multiply. 26 25
5 18
a ab
b
Solution 26 25
5 18
a ab
b
2 3 5 5
5 2 3 3
a a b b
b
5
1 3
a a b
25
3
a b
Slide 7- 24Copyright © 2011 Pearson Education, Inc.
Example 2a
Solution
1 1
2 x x
Write numerators and denominators in factored form.
Multiply the remaining numerator factors and denominator factors.
3
5 3 12
2 8 15
x x
x x
3 45
2 4 5 3
xx
x x x x
2
1
2x
Slide 7- 25Copyright © 2011 Pearson Education, Inc.
Example 2b
Multiply. 220 4 3
9 2 10
x x
x x
Solution
2 1
3 1
x
Write numerators and denominators in factored form.
Multiply the remaining numerator factors and denominator factors.
4 5 3
3 3 2 5
x x x
x x
2 2 1 5 3
3 3 2 5
x x x
x x
2
3
x
Slide 7- 26Copyright © 2011 Pearson Education, Inc.
2 1 2 2
2 2 5 2 1 1
x x
x x
Example 2c
Multiply. 2
2
2 4
20 2 3 1
x x
x x
Solution2
2
2 4
20 2 3 1
x x
x x
2 2
or 5 2 1 10 5
x x
x x
Write numerators and denominators in factored form.
Multiply the remaining numerator factors and denominator factors.
Slide 7- 27Copyright © 2011 Pearson Education, Inc.
Dividing Rational Expressions
, where , , , and
are polynomials and 0, 0, and 0.
P R P SP Q R S
Q S Q R
Q R S
Slide 7- 28Copyright © 2011 Pearson Education, Inc.
Dividing Rational ExpressionsTo divide rational expressions,
1. Write an equivalent multiplication statement with the reciprocal of the divisor.
2. Factor each numerator and denominator completely. (Steps 1 and 2 are interchangeable.)
3. Divide out any numerator factor with any matching denominator factor.
4. Multiply numerator by numerator and denominator by denominator.
5. Simplify as needed.
Slide 7- 29Copyright © 2011 Pearson Education, Inc.
Example 3
Divide . 2
3
6 20
35 5
x y y
z z
Solution
2
2
3
70
x
z
Write an equivalent multiplication statement.
Divide out common factors, and multiply remaining factors.
2
3
6 20
35 5
x y y
z z
2
3
6 5
35 20
x y z
z y
3 2 5
5 7 5 2 2
x x y z
z z z y
Slide 7- 30Copyright © 2011 Pearson Education, Inc.
2 1 3 3 1 2
2 1 2 3 1 2 1
x x x x
x x x x
Example 4
Divide. 2 2
2 2
2 7 3 6 5 1
2 3 2 3 5 2
x x x x
x x x x
Solution 2 2
2 2
2 7 3 6 5 1
2 3 2 3 5 2
x x x x
x x x x
2 2
2 2
2 7 3 3 5 2
2 3 2 6 5 1
x x x x
x x x x
3
2 1
x
x
Write an equivalent multiplication statement.
Divide out common factors, and multiply remaining factors.
Slide 7- 31Copyright © 2011 Pearson Education, Inc.
Using Dimensional Analysis to Convert Between Units of MeasurementTo convert units using dimensional analysis, multiply the given measurement by conversion factors so that the undesired units divide out, leaving the desired units.
Slide 7- 32Copyright © 2011 Pearson Education, Inc.
Example 5a
Convert 300 ounces to pounds
Solution
300 oz. 1 lb.
1 16 oz. 18.75 lb.300 oz.
300 lb.
16
Slide 7- 33Copyright © 2011 Pearson Education, Inc.
Example 5b
Convert 17 yards to inches.
Solution
17 yd. 3 ft. 12 in.
1 1 yd. 1 ft. 612 inches17 yards
Slide 7- 34Copyright © 2011 Pearson Education, Inc.
Example 5c
Convert 25 miles per hour to feet per second.
Solution
25 mi. 5280ft. 1 hr. 1min .
1 hr. 1 mi. 60 min. 60sec.
25 miles
1 hour36.6 ft./sec.
Slide 7- 35Copyright © 2011 Pearson Education, Inc.
Multiply.
a)
b)
c)
d)
2
2
6 9 2
4 3
m m m
m m
3
2
m
m
3
2
m
m
3
2
m
m
3
2
m
m
7.2
Slide 7- 36Copyright © 2011 Pearson Education, Inc.
Multiply.
a)
b)
c)
d)
2
2
6 9 2
4 3
m m m
m m
3
2
m
m
3
2
m
m
3
2
m
m
3
2
m
m
7.2
Slide 7- 37Copyright © 2011 Pearson Education, Inc.
Divide.
a)
b)
c)
d)
2 5 6 2
6 6
n n n
n n
3n
3
2
n
n
3 2
2
n n
n
3 2
2
n n
n
7.2
Slide 7- 38Copyright © 2011 Pearson Education, Inc.
Divide.
a)
b)
c)
d)
2 5 6 2
6 6
n n n
n n
3n
3
2
n
n
3 2
2
n n
n
3 2
2
n n
n
7.2
Copyright © 2011 Pearson Education, Inc.
Adding and Subtracting Rational Expressions with the Same Denominator7.37.3
1. Add or subtract rational expressions with the same denominator.
Slide 7- 40Copyright © 2011 Pearson Education, Inc.
Adding or Subtracting Rational Expressions (Same Denominator)To add or subtract rational expressions that have the same denominator:1. Add or subtract the numerators and keep the same denominator. 2. Simplify to lowest terms. (Remember to factor the
numerators and denominators completely in order to simplify).
Slide 7- 41Copyright © 2011 Pearson Education, Inc.
Example 1
Add.
Solution
7
18 18
y y
7
18 18
y y 7
18
y y
8
18
y
2 2 2
3 3 2
y
4
9
y
Since the rational expressions have the same denominator, we add numerators and keep the same denominator.
Factor.
Divide out the common factor, 2.
Slide 7- 42Copyright © 2011 Pearson Education, Inc.
Example 2a
Subtract.
Solution
18
7 7
z
z z
Warning: Although it may be tempting to do so, we cannot divide out the z’s because they are terms, not factors.
18
7
z
z
18
7 7
z
z z
Slide 7- 43Copyright © 2011 Pearson Education, Inc.
Example 2b
Subtract.
Solution
2 16
4 4
x
x x
4 4
4
x x
x
2 16
4 4
x
x x
2 16
4
x
x
4x Divide out the common factor, x – 4.
Note: The numerator can be factored, so we may be able to simplify.
Slide 7- 44Copyright © 2011 Pearson Education, Inc.
Example 2c
Subtract.
Solution
7 11
12 12
x
7 11
12
x
7 11
12 12
x
7 11
12
x
Slide 7- 45Copyright © 2011 Pearson Education, Inc.
2 3
5
b b
b b
Example 3
Add.
Solution
2 2
2 2
7 4
5 5
b b b b
b b b b
Combine like terms in the numerator.
2 2
2 2
7 4
5 5
b b b b
b b b b
2 2
2
7 4
5
b b b b
b b
2
2
2 3
5
b b
b b
2 3
5
b
b
Factor the numerator and the denominator.
Divide out the common factor, b.
Slide 7- 46Copyright © 2011 Pearson Education, Inc.
Example 4
Add.
Solution
2 2
3 3
5 6 3 2
2 8 2 8
x x x x
x x x x
Combine like terms in the numerator.
Factor the numerator and the denominator.
2 2
3 3
5 6 3 2
2 8 2 8
x x x x
x x x x
2 2
3
5 6 3 2
2 8
x x x x
x x
2
3
2 8 8
2 8
x x
x x
2 2 2
2 2 2
x x
x x x
Slide 7- 47Copyright © 2011 Pearson Education, Inc.
continued
2
2 2 or
2 2
x x
x x x x
2 2 2
2 2 2
x x
x x x
Divide out the common factors, 2 and x + 2.
Slide 7- 48Copyright © 2011 Pearson Education, Inc.
Example 5a
Subtract.
Solution
2 23 2 7 8
3 1 3 1
x x x x
x x
2 23 2 7 8
3 1
x x x x
x
22 1
3 1
x x
x
2 23 2 7 8
3 1 3 1
x x x x
x x
2 23 2 7 8
3 1
x x x x
x
Note: To write an equivalent addition, change the operation symbol from a minus sign to a plus sign and change all the signs in the subtrahend (second) polynomial.
Slide 7- 49Copyright © 2011 Pearson Education, Inc.
Example 5b
Subtract.
Solution
3
2 2
5 5 5 22
2 6 2 6
x x x
x x x x
3
2
5 5 5 22
2 6
x x x
x x
Note: This numerator is the difference of cubes. 3
2
5 5 5 22
2 6
x x x
x x
3
2
27
2 6
x
x x
3
2 2
5 5 5 22
2 6 2 6
x x x
x x x x
Slide 7- 50Copyright © 2011 Pearson Education, Inc.
continued
Note: This numerator is the difference of cubes.
23 3 9
2 3
x x x
x x
3
2
27
2 6
x
x x
2 3 9
2
x x
x
Slide 7- 51Copyright © 2011 Pearson Education, Inc.
Add.
a)
b)
c)
d)
2 2
2 5 7
3 4 3 4
x x
x x x x
2
3 12
3 4
x
x x
3 4
1 4
x
x x
2
1 4
x
x x
3
1x
7.3
Slide 7- 52Copyright © 2011 Pearson Education, Inc.
Add.
a)
b)
c)
d)
2 2
2 5 7
3 4 3 4
x x
x x x x
2
3 12
3 4
x
x x
3 4
1 4
x
x x
2
1 4
x
x x
3
1x
7.3
Slide 7- 53Copyright © 2011 Pearson Education, Inc.
Subtract.
a)
b)
c)
d)
2 2
1 5 3
2 1 2 1
a a
a a a a
2
2 6
1
a
a
2
2 3
1
a
a
2
4
1a
4
1a
7.3
Slide 7- 54Copyright © 2011 Pearson Education, Inc.
Subtract.
a)
b)
c)
d)
2 2
1 5 3
2 1 2 1
a a
a a a a
2
2 6
1
a
a
2
2 3
1
a
a
2
4
1a
4
1a
7.3
Copyright © 2011 Pearson Education, Inc.
Adding and Subtracting Rational Expressions with Different Denominators7.47.4
1. Find the LCD of two or more rational expressions.2. Given two rational expressions, write equivalent rational
expressions with their LCD.3. Add or subtract rational expressions with different
denominators.
Slide 7- 56Copyright © 2011 Pearson Education, Inc.
Remember that when adding or subtracting fractions with different denominators, we must first find a common denominator. It is helpful to use the least common denominator (LCD), which is the smallest number that is evenly divisible by all the denominators.
38 2 212 2 3
3LCD 2 3 = 24
Slide 7- 57Copyright © 2011 Pearson Education, Inc.
Finding the LCDTo find the LCD of two or more rational expressions,1. Find the prime factorization of each denominator. 2. Write a product that contains each unique prime
factor the greatest number of times that factor occurs in any factorization. Or if you prefer to use exponents, write the product that contains each unique prime factor raised to the greatest exponent that occurs on that factor in any factorization.
3. Simplify the product found in step 2.
Slide 7- 58Copyright © 2011 Pearson Education, Inc.
Example 1
Find the LCD of
Solution We first factor the denominators 12y2 and 8y3 by writing their prime factorizations.
2 3
5 2 and .
12 8y y
2 2 212 2 3 y y 3 3 38 2 y y
The unique factors are 2, 3, and y. To generate the LCD, include 2, 3, and y the greatest number of times each appears in any of the factorizations.
Slide 7- 59Copyright © 2011 Pearson Education, Inc.
continued
Note: We can compare exponents in the prime factorizations to create the LCD. If two factorizations have the same prime factors, we write that prime factor in the LCD with the greater of the two exponents.
3 3 3LCD = 2 3 24y y
The greatest number of times that 2 appears is three times (in 23 • y3).
The greatest number of times that 3 appears is once (in 22 • 3 • y2).
The greatest number of times that y appears is three times (in 23 • y3).
Slide 7- 60Copyright © 2011 Pearson Education, Inc.
Example 2
Find the LCD.
Solution Factor the denominators x2 – 25 and 2x – 10.
2
8 3 and
25 2 10x x
2 25 5 5x x x 2 10 2 5x x
The unique factors are 2, (x + 5), and (x – 5). The greatest number of times that 2 appears is once.The greatest number of times that (x + 5) appears is once.The greatest number of times that (x – 5) appears is once.
LCD = 2 5 5x x
Slide 7- 61Copyright © 2011 Pearson Education, Inc.
Example 3
Write as equivalent rational expressions with their LCD.
Solution The LCD is 24x2. For each rational expression, we multiply both the numerator and denominator by an appropriate factor so the denominator becomes 24x2.
2
7 5 and
8 12x x
7
8x
7 3
8 3
x
x x
2
21
24
x
x
2
5
12x 2
5 2
12 2x
2
10
24x
Slide 7- 62Copyright © 2011 Pearson Education, Inc.
Example 4
Write as equivalent rational expressions with their LCD.
Solution In the previous example, we found the LCD to be .
2
8 3 and
25 2 10x x
2 5 5x x
2
8
25x 8
5 5x x
8 2
5 5 2x x
16
2 5 5x x
Write the denominator in factored form.
Note: Another way to determine the appropriate factor is to think of it as the factor in the LCD that is missing from the original denominator.
Multiply the numerator and the denominator by the same factor, 2, to get the LCD, 2(x + 5)(x – 5).
Slide 7- 63Copyright © 2011 Pearson Education, Inc.
continued
3
2 10x 3
2 5x
3 5
2 5 5
x
x x
3 5
2 5 5
x
x x
Write the denominator in factored form.
Multiply the numerator and the denominator by the same factor, (x + 5), to get the LCD, 2(x + 5)(x – 5).
Slide 7- 64Copyright © 2011 Pearson Education, Inc.
Adding or Subtracting Rational Expressions with Different DenominatorsTo add or subtract rational expressions with different denominators,1. Find the LCD. 2. Write each rational expression as an equivalent expression with the LCD.3. Add or subtract the numerators and keep the LCD.4. Simplify.
Slide 7- 65Copyright © 2011 Pearson Education, Inc.
Example 5
Add.
Solution The LCD is 24x2.
2
3 5 1 +
8 6
x x
x x
2
3 5 1 +
8 6
x x
x x
2
3 3 5 1 4 +
8 3 6 4
x x x
x x x
Write equivalent rational
expressions with the LCD, 24x2.
2
2 2
3 9 20 4 +
24 24
x x x
x x
2
2
3 9 20 4
24
x x x
x
2
2
3 29 4
24
x x
x
Add numerators.
Note: Remember that to add polynomials, we combine like terms.
Slide 7- 66Copyright © 2011 Pearson Education, Inc.
Example 6
Add .
Solution
2 2
6 4 +
2 4 3x x x x
2 2
6 4 +
2 4 3x x x x
6 4
+ 2 1 3 1x x x x
6 3 4 2 +
2 1 3 2 3 1
x x
x x x x x x
Slide 7- 67Copyright © 2011 Pearson Education, Inc.
continued
6 3 4 2 +
2 1 3 2 3 1
x x
x x x x x x
6 18 4 8
+ 2 1 3 2 3 1
x x
x x x x x x
6 18 4 8
2 1 3
x x
x x x
10 10
2 1 3
x
x x x
10 1
2 1 3
x
x x x
10
2 3x x
Slide 7- 68Copyright © 2011 Pearson Education, Inc.
Example 7
Add.
Solution
7 3 +
6 6
x
x x
Since x – 6 and 6 – x are additive inverses, we obtain the LCD by multiplying the numerator and denominator of one of the rational expressions by –1. We chose the second rational expression.
7 3 +
6 6
x
x x
3 17
+ 6 6 1
x
x x
7 3 +
6 6
x
x x
3 7
6
x
x
Slide 7- 69Copyright © 2011 Pearson Education, Inc.
Add.
a)
b)
c)
d)
5 8
4 3 12n n
23
3 12n 13
3 12n
13 4
3 12
n
n
5
3 12n
7.4
Slide 7- 70Copyright © 2011 Pearson Education, Inc.
Add.
a)
b)
c)
d)
5 8
4 3 12n n
23
3 12n 13
3 12n
13 4
3 12
n
n
5
3 12n
7.4
Slide 7- 71Copyright © 2011 Pearson Education, Inc.
Subtract.
a)
b)
c)
d)
5 2 3
5 5
x x
x x
3 1
5
x
x
7 3
5
x
x
7 3
5
x
x
7 3
5
x
x
7.4
Slide 7- 72Copyright © 2011 Pearson Education, Inc.
Subtract.
a)
b)
c)
d)
5 2 3
5 5
x x
x x
3 1
5
x
x
7 3
5
x
x
7 3
5
x
x
7 3
5
x
x
7.4
Copyright © 2011 Pearson Education, Inc.
Complex Rational Expressions7.57.5
1. Simplify complex rational expressions.
Slide 7- 74Copyright © 2011 Pearson Education, Inc.
Complex rational expression: A rational expression that contains rational expressions in the numerator or denominator.
Examples:
3456
3
2
xx
y7
2tt
2 14
13
nn
n
Slide 7- 75Copyright © 2011 Pearson Education, Inc.
Simplifying Complex Rational ExpressionsTo simplify a complex rational expression, use one of the following methods.Method 11. Simplify the numerator and denominator if needed.2. Rewrite as a horizontal division problem. Method 21. Multiply the numerator and denominator of the complex rational expression by their LCD. 2. Simplify.
Slide 7- 76Copyright © 2011 Pearson Education, Inc.
Example 1 —Method 1
Simplify.
Solution Write the numerator fractions as equivalent fractions with their LCD, 12, and write the denominator fractions with their LCD, 24.
3 54 65 18 6
3 54 65 18 6
3(3) 5(2)4(3) 6(2)5(3) 1(4)8(3) 6(4)
9 1012 1215 424 24
19121124
Slide 7- 77Copyright © 2011 Pearson Education, Inc.
continued
19 11
12 24
19 24
12 11
38
11
1
2
19121124
Slide 7- 78Copyright © 2011 Pearson Education, Inc.
Example 1—Method 2
Simplify.
Solution Multiply the numerator and denominator by 24.
3 54 65 18 6
3 54 65 18 6
243 54 65 1
26
48
3 24 5 244 1 6 15 24 1 248 1 6 1
1
46
13
1
4
1
18 20
15 4
38
11
Slide 7- 79Copyright © 2011 Pearson Education, Inc.
Example 1 Simplify.
Solution Write the numerator and denominator as equivalent fractions. (Method 1)
12
23
y
y
12
23
y
y
2( ) 1(1)1( ) (1)3( ) 2(1)1( ) (1)
yy yyy y
2 1
3 2
yy
yy
2 1 3 2
y y
y y
2 1
3 2
y y
y y2 1
3 2
y
y
2 1
3 2
yy yyy y
Slide 7- 80Copyright © 2011 Pearson Education, Inc.
Example 1Simplify.
Solution Multiply the numerator and denominator by the LCD of all the rational expressions. (Method 2)
32
11
xx
xx
32
11
xx
xx
32
11
2
2
xx
xx
x
x
2 3 22 1 1
1 2 1 21 1 1
x x xx
x x xx
2 6
2 2( 1)
x
x x
2 6
2 2 2
x
x x
2 26 6 or
4 2 2(2 1)
x x
x x
Slide 7- 81Copyright © 2011 Pearson Education, Inc.
Simplify.
a)
b)
c)
d)
abcd
ab
cd
ac
bdad
bc
1ad
bc
7.5
Slide 7- 82Copyright © 2011 Pearson Education, Inc.
Simplify.
a)
b)
c)
d)
abcd
ab
cd
ac
bdad
bc
1ad
bc
7.5
Slide 7- 83Copyright © 2011 Pearson Education, Inc.
Simplify.
a)
b)
c) a + 1
d) a – 1
1
11
aa
a
1a
a
1a
a
7.5
Slide 7- 84Copyright © 2011 Pearson Education, Inc.
Simplify.
a)
b)
c) a + 1
d) a – 1
1
11
aa
a
1a
a
1a
a
7.5
Copyright © 2011 Pearson Education, Inc.
Solving Equations Containing Rational Expressions7.67.6
1. Solve equations containing rational expressions.
Slide 7- 86Copyright © 2011 Pearson Education, Inc.
Example 1Solve.
Solution
3
6 4 8
x x
3
6 4 8
x x
24 243
6 4 8
x x
324 2
62
84
44
x x
4 6 9 x x2 9 x2 9
2 2
x
9
2x
Multiply both sides by 24.
Distribute and divide out common factors.
Subtract 6x from both sides.
Divide both sides by 2.
Simplify both sides.
4
1 1
6
1
3
Slide 7- 87Copyright © 2011 Pearson Education, Inc.
Example 2Solve.
Solution
2 5 1
4x x
4 42 5 1
4x x
x x
Multiply both sides by 4x.
Distribute and divide out common factors.
Simplify both sides.
2 5 1
4x x
42 5 1
44 4
x xx x x
8 20 x
12x
Slide 7- 88Copyright © 2011 Pearson Education, Inc.
Extraneous Solution: An apparent solution that does not solve its equation.
Solving Equations Containing Rational Expressions To solve an equation that contains rational expressions, 1. Eliminate the rational expressions by multiplying both sides of the equation by their LCD. 2. Solve the equation using the methods we learned in Chapters 2 (linear equations) and 7 (quadratic equations).3. Check your solution(s) in the original equation. Discard any extraneous solutions.
Slide 7- 89Copyright © 2011 Pearson Education, Inc.
Example 3Solve.
Solution If x = 6, then the equation is undefined, so the solution cannot be 6.
2
6 6
x
x
2
6 6
x
x
6( 6) 6( 6)2
6 6
xx
xx
6 ( 6)2 x x
6 2 12 x x
4 12x
3x
Divide out common factors.
Distribute 2 to clear the parentheses.
Subtract 2x on both sides.
Divide both sides by 4.
Slide 7- 90Copyright © 2011 Pearson Education, Inc.
Example 4
Solve.Solution
2 1 6
3 1 3 1
x
x x x
(3 1) (2 1 6
3 33
1 11)
x x xx
x x xx
(3 1) (3 1) (2 1 6
3 1 3 13 1)
x
x xx x x x
xx x
2 3 1 6 ( ) x x x x22 3 1 6 x x x
26 1 0x x (3 1)(2 1) 0 x x
3 1 0 or 2 1 0 x x 1 1 or
3 2 x x
In the original equation, when x = -1/3 it is undefined, so the only solution is x = ½.
Slide 7- 91Copyright © 2011 Pearson Education, Inc.
Example 5
Solve.
Solution
2
2 11
7 10 21 3
a a
a a a a
2 11
7 ( 7)( 3) 3
a a
a a a a
2 1( 7)( 3) 1 ( 7)( 3)
7 ( 7)( 3) 3
a aa a a a
a a a a
2 ( 3) ( 7)( 3) 1 ( 7) a a a a a a
2 2 22 6 ( 10 21) 1 7 a a a a a a
2 2 22 6 10 21 1 7 a a a a a a2 24 21 7 1 a a a a
Slide 7- 92Copyright © 2011 Pearson Education, Inc.
continued
2 24 21 7 1 a a a a
11 21 1 a
11 22 a
11 22
11 11
a
2a
2
2 11 .
7 10 21 3
a a
a a a a
This solves the original equation
Slide 7- 93Copyright © 2011 Pearson Education, Inc.
Solve.
a) 2
b) 5
c) 7
d) 10
5 7
20 2
a
a
7.6
Slide 7- 94Copyright © 2011 Pearson Education, Inc.
Solve.
a) 2
b) 5
c) 7
d) 10
5 7
20 2
a
a
7.6
Slide 7- 95Copyright © 2011 Pearson Education, Inc.
Solve.
a) 2
b) 1
c) 2 and 5
d) 3 and 4
2
2 10 1 2 10
1 1 1
a a a
a a a
7.6
Slide 7- 96Copyright © 2011 Pearson Education, Inc.
Solve.
a) 2
b) 1
c) 2 and 5
d) 3 and 4
2
2 10 1 2 10
1 1 1
a a a
a a a
7.6
Copyright © 2011 Pearson Education, Inc.
Applications with Rational Expressions, Including Variation7.77.7
1. Use tables to solve problems with two unknowns involving rational expressions.
2. Solve problems involving direct variation.3. Solve problems involving inverse variation.4. Solve problems involving joint variation.5. Solve problems involving combined variation.
Slide 7- 98Copyright © 2011 Pearson Education, Inc.
Tables are helpful in problems involving two or more people working together to complete a task.For each person involved, the rate of work, time at work, and amount of the task completed are related as follows:
Since the people are working together, the sum of their individual amounts of the task completed equals the whole task.
Person’s rate of work
Time at work
Amount of the task completed by that person
=
Amount completed by one person
Amount completed by the other person
Whole task+ =
Slide 7- 99Copyright © 2011 Pearson Education, Inc.
Example 1Louis and Rebecca own a painting business. Louis can paint an average size room in 5 hours. Rebecca can paint the same room in 3 hours. How long would it take them to paint the same room working together?
UnderstandLouis paints at a rate of 1 room in 5 hours, or 1/5 of a room per hour.Rebecca paints at a rate of 1 room in 3 hours, or 1/3 of a room per hour.
Slide 7- 100Copyright © 2011 Pearson Education, Inc.
continued
Plan and Execute Louis’ amount completed + Rebecca’s amount completed = 1 room
Category Rate of Work
Time at Work
Amount of Task Completed
Louis t t
Rebecca t t
1513
1513
1 11
5 3t t
Slide 7- 101Copyright © 2011 Pearson Education, Inc.
continued 1 1
15 3
t t
1 115 15(1)
5 3t t
1 1
15 15 155 3
t t
3 5 15t t 8 15t
15
8t
3
1
5
1
Slide 7- 102Copyright © 2011 Pearson Education, Inc.
continued
AnswerWorking together, it takes Louis and Rebecca or 1 7/8 hours to paint an average size room.
Check
158
1 15 1 151
5 8 3 8
15 151
40 24
45 751
120 120
1201
120
Slide 7- 103Copyright © 2011 Pearson Education, Inc.
Example 2A sports car travels 15 mph faster than a loaded truck on the freeway. In the same time that the sports car travels 156 miles, the truck travels 120 miles. Find the speed of each vehicle.
UnderstandUse a table to organize the information.
Vehicle Distance Rate Time
Sports car 156 miles r + 15
Truck 120 miles r
156
15r 120
r
Slide 7- 104Copyright © 2011 Pearson Education, Inc.
continuedPlan and Execute
156 12015 15
15r r r r
r r
156 120
15r r
156 120 15r r
156 120 1800r r
36 1800r
50r
Slide 7- 105Copyright © 2011 Pearson Education, Inc.
continued
AnswerThe truck travels at 50 mph while the sports car travels at r + 15 or 65 mph.
Check 156 120
15r r
156 120
65 50
2.4 2.4
Slide 7- 106Copyright © 2011 Pearson Education, Inc.
Direct variation: Two variables, y and x, are in direct variation if y = kx, where k is a constant.
In words, direct variation is written as “y varies directly as x” or “y is directly proportional to x” and these phrases translate to y = kx.
Slide 7- 107Copyright © 2011 Pearson Education, Inc.
Example 3Suppose y varies directly as x. When y = 12, x = 6. Find y when x = 9.Solution Replace y with 12 and x with 6, then solve for k.
y = kx12 = k 62 = k
Replace k with 2 in y = kx so that we have y = 2x.y = 2xy = 2(9) = 18
Slide 7- 108Copyright © 2011 Pearson Education, Inc.
Example 4Nigel’s paycheck varies directly as the number of hours worked. For 15 hours of work, the pay is $188.25. Find the pay for 27 hours of work.Understand Translating “paycheck varies directly as the number of hours worked,” we write p = kh, where p represents the paycheck and h represents the hours.Plan Use p = kh and replace p with 188.25 and h with 15, in order to find the value of k.
Slide 7- 109Copyright © 2011 Pearson Education, Inc.
continuedExecute 188.25 = k 15
12.55 = k
Replace k with 12.55. p = kh p = 12.55h
When hours = 27: p = 12.55(27) p = 338.85
Answer Working 27 hours will earn a paycheck of $338.85.
Slide 7- 110Copyright © 2011 Pearson Education, Inc.
Inverse variation: Two variables, y and x, are in inverse variation if where k is a constant.
In words, inverse variation is written as “y varies inversely as x” or “y is inversely proportional to x, and these phrases translate to .”
,k
yx
,k
yx
Slide 7- 111Copyright © 2011 Pearson Education, Inc.
Example 5If the temperature is constant, the pressure of a gas in a container varies inversely as the volume of the container. If the pressure is 15 pounds per square foot in a container with 4 cubic feet, what is the pressure in a container with 1.5 cubic feet?Understand Because the pressure and volume vary inversely, we can write
where P represents the pressure and V represents the volume.
kP
V
Slide 7- 112Copyright © 2011 Pearson Education, Inc.
continuedPlan Use pressure is 15 when volume is 4 to determine the value of k.Execute
Answer The pressure is 40 pounds per square foot when the volume is 1.5 cubic feet.
kP
V
154
k
60 k
Replacing k with 60, solve for P when V is 1.5.
60P
V
60
1.5P 40
Slide 7- 113Copyright © 2011 Pearson Education, Inc.
Joint variation: If y varies jointly as x and z, then y = kxz, where k is the constant of variation.
Slide 7- 114Copyright © 2011 Pearson Education, Inc.
Solution
Suppose a varies jointly with b and c. If a = 72 when b = 3 and c = 8, find a when b = 4 and c = 2.
We have a = kbc, so
72 3 8k
3 kThe equation of variation is a = 3bc.
a = 3bc
a = 3(4)(2) = 24
Example 6
Slide 7- 115Copyright © 2011 Pearson Education, Inc.
Example 7 The volume of wood V in a tree varies jointly as the height h and the square of the girth g (girth is the distance around). The volume of a redwood tree is 216 m3 when the height is 30 m and the girth is 1.5 m, what is the height of a tree whose volume is 1700 m3 and girth is 2.5 m?Solution Find k using the first set of data.
V = khg2
216 = k 30 1.52
3.2 = k
Slide 7- 116Copyright © 2011 Pearson Education, Inc.
continued
The equation of variation is V = 3.2hg2. We substitute the second set of data into the equation.Volume is 1700 m3 and girth is 2.5 m.
V = 3.2hg2
1700 = 3.2 h 2.52
1700 = 20h85 = h
The height of the tree is 85 m.
Slide 7- 117Copyright © 2011 Pearson Education, Inc.
Danielle can paint a 4800-square foot house in 8 days, and Paige can do the same job in only 6 days. How long will it take them to paint a 4800-square foot house if they work together?
a) 14 days
b) 7 days
c)
d)
33 days
72
3 days5
7.7
Slide 7- 118Copyright © 2011 Pearson Education, Inc.
Danielle can paint a 4800-square foot house in 8 days, and Paige can do the same job in only 6 days. How long will it take them to paint a 4800-square foot house if they work together?
a) 14 days
b) 7 days
c)
d)
33 days
72
3 days5
7.7
Slide 7- 119Copyright © 2011 Pearson Education, Inc.
The distance a car travels varies directly with the amount of gas it carries. On the highway, a van travels 112 miles using 7 gallons of gas. How many gallons are required to travel 544 miles?
a) 15 gallons
b) 16 gallons
c) 32 gallons
d) 34 gallons
7.7
Slide 7- 120Copyright © 2011 Pearson Education, Inc.
The distance a car travels varies directly with the amount of gas it carries. On the highway, a van travels 112 miles using 7 gallons of gas. How many gallons are required to travel 544 miles?
a) 15 gallons
b) 16 gallons
c) 32 gallons
d) 34 gallons
7.7
Slide 7- 121Copyright © 2011 Pearson Education, Inc.
If y varies inversely as x and y = 8 when x = 10, what is y when x is 16?
a) 2
b) 5
c) 10
d) 16
7.7
Slide 7- 122Copyright © 2011 Pearson Education, Inc.
If y varies inversely as x and y = 8 when x = 10, what is y when x is 16?
a) 2
b) 5
c) 10
d) 16
7.7